where did this come from.

[allegansveritatem]

Here is an example problem that contains an expression I don't get and it is not at all introduced in the text. Here is the example:

My problem with this is I don't see anything about C being a tangent anywhere in the original layout of the equation. Am I supposed to know that a variable in that position is by some definition a tangent? So, whee did the tangent come from. I seem to vaguely recall discovering on my own that the phase shift was a tangent of something...but I don't know if that applies here.

 

[Steven G]

Where does it say that C is a tangent?

Do you have a question about the proof? Or is your question about C being tangent to something? If your question is about not understanding something in the proof then please tell us where you are getting lost?

Please post back

 

[Dr.Peterson]

I'm confused. The problem states explicitly that tan(C) = b/a; but that means C is an angle, not a tangent.

Or are you asking how they would come up with C when trying to rewrite the expression a cos(Bx) + b sin(Bx)?

 

[lev888]

It's in the line beginning with "where".

 

[Steven G]

Dr Peterson, you really are slowing down. I am starting to worry about you.

 

[Steven G]

It's in the line beginning with "where".

Those are just definitions which you should accept.

 

[lev888]

Those are just definitions which you should accept.

Yes. And I'm guessing OP missed it.

 

[Dr.Peterson]

I've seen many ways to prove this fact, some clearer than others; I have to say that this one seems particularly poorly motivated, pulling that tangent out of a hat. WHY would you choose to let C have that value?

I would instead have done something like this:

We have the expression [MATH]a\cos(Bx) + b\sin(Bx)[/MATH] . That is vaguely reminiscent of the angle sum formula, [MATH]\cos(u - v) = \cos(u)\cos(v) + \sin(u)\sin(v)[/MATH].

We'd like to turn [MATH]a[/MATH] into a cosine and [MATH]b[/MATH] into a sine. But that would require [MATH]a^2 + b^2 = 1[/MATH]. We can move in that direction by dividing everything by [MATH]\sqrt{a^2 + b^2}[/MATH], and then multiplying by it:

[MATH]a\cos(Bx) + b\sin(Bx) = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\cos(Bx) + \frac{b}{\sqrt{a^2 + b^2}}\sin(Bx)\right)[/MATH]​

Now we can call [MATH]\sqrt{a^2 + b^2}[/MATH] "[MATH]A[/MATH]", and suppose that [MATH]\frac{a}{\sqrt{a^2 + b^2}} = \cos(C)[/MATH] and [MATH]\frac{b}{\sqrt{a^2 + b^2}} = \sin(C)[/MATH], and everything will work out.

But do you see that then, [MATH]\tan(C) = \frac{b}{a}[/MATH]?

 

[topsquark]

Dr Peterson, you really are slowing down. I am starting to worry about you.

Tsk tsk. Don't doubt the master!

-Dan

 

[Steven G]

Tsk tsk. Don't doubt the master!

-Dan

I would never doubt Dr Peterson, but I am concerned about him slowing down a bit. Maybe he taught me well and I am speeding up?

 

[Dr.Peterson]

The truth is, I was pondering whether to give the long answer immediately, but when I saw someone else had answered, I posted the short answer I'd written so far, just to not look like I was responding to whatever you had written. Then I decided the full answer would be appropriate, and continued with what I was doing.

So in one sense you sped me up, and in another sense, I took longer because I thought more ...

 

[Steven G]

The truth is, I was pondering whether to give the long answer immediately, but when I saw someone else had answered, I posted the short answer I'd written so far, just to not look like I was responding to whatever you had written. Then I decided the full answer would be appropriate, and continued with what I was doing.

So in one sense you sped me up, and in another sense, I took longer because I thought more ...

...and I thought that I was getting better. Oh well. The main thing is that in the end I really have learned to think better from you and that is so valuable. Thank you!

 

[allegansveritatem]

I'm confused. The problem states explicitly that tan(C) = b/a; but that means C is an angle, not a tangent.

Or are you asking how they would come up with C when trying to rewrite the expression a cos(Bx) + b sin(Bx)?

what I mean is: the statement of the problem, the equation to be solved which is: a cosBx + bsinBx=Acos(Bx-C). There is no mention of C being a tanC. Later down in the working out of the example the author starts refering to C as a tanC. I want to know where he got that tan? I mean, is it understood that C in the position it is in here in the phase shift position, is always a tan and so there is no need to say Acos(Bx-tanC)?

 

[Steven G]

Can you state exactly where you feel that C became tanC?

 

[allegansveritatem]

I've seen many ways to prove this fact, some clearer than others; I have to say that this one seems particularly poorly motivated, pulling that tangent out of a hat. WHY would you choose to let C have that value?

I would instead have done something like this:

We have the expression [MATH]a\cos(Bx) + b\sin(Bx)[/MATH] . That is vaguely reminiscent of the angle sum formula, [MATH]\cos(u - v) = \cos(u)\cos(v) + \sin(u)\sin(v)[/MATH].

We'd like to turn [MATH]a[/MATH] into a cosine and [MATH]b[/MATH] into a sine. But that would require [MATH]a^2 + b^2 = 1[/MATH]. We can move in that direction by dividing everything by [MATH]\sqrt{a^2 + b^2}[/MATH], and then multiplying by it:

[MATH]a\cos(Bx) + b\sin(Bx) = \sqrt{a^2 + b^2}\left(\frac{a}{\sqrt{a^2 + b^2}}\cos(Bx) + \frac{b}{\sqrt{a^2 + b^2}}\sin(Bx)\right)[/MATH]​

Now we can call [MATH]\sqrt{a^2 + b^2}[/MATH] "[MATH]A[/MATH]", and suppose that [MATH]\frac{a}{\sqrt{a^2 + b^2}} = \cos(C)[/MATH] and [MATH]\frac{b}{\sqrt{a^2 + b^2}} = \sin(C)[/MATH], and everything will work out.

But do you see that then, [MATH]\tan(C) = \frac{b}{a}[/MATH]?

I will have to go over this post in the morning when my head is screwed back on. But let me me say that I can recall a few weeks ago playing with some of these trig functions and discovering that the phase shift was always a tangent,i.e., the opp over the adjacent--but right now I can't quite remember how I knew what the opp was and what the adjacent was.All I remember is the phase shift was the same as the tangent. The book never mentioned the fact but I found it to be so.

 

[allegansveritatem]

Can you state exactly where you feel that C became tanC?

well, it was not stated in the equation on top that C was tanC. Then, in the working out of the problem they introduced C as tanC. Why?

 

[allegansveritatem]

Those are just definitions which you should accept.

so you are saying the tan is arbitrary here and not to be understood as necessarily so given the position of C?

 

[Steven G]

well, it was not stated in the equation on top that C was tanC. Then, in the working out of the problem they introduced C as tanC. Why?

Please state where they introduced C as tanC?

I promise you that they did not do that. if you could please tell me where you think that made this change I will show you what happened. You do remember that they said that a and b are connected by tan(C) = b/a. So then b=a tan(C). Then theny can replace any and all b's with tan(C). This is how tan(C) came into the equation. Never was C replaced with tan(C).

 

[lev888]

author starts refering to C as a tanC

No, it's b=a*tanC. Per problem statement, in the line beginning with "where", as I already noted.

 

[Dr.Peterson]

well, it was not stated in the equation on top that C was tanC. Then, in the working out of the problem they introduced C as tanC. Why?

I'm presuming that this is an April Fool's joke, because I know you're more intelligent than this.

No one ever said that "C was tan C"; the number C is not the same as the tangent of C. And the tangent function was not introduced in the solution, but stated in the problem: "where [MATH]\tan C = \frac{b}{a}[/MATH]. That is, part of what you are to prove is that if you define C as an angle whose tangent is b/a, then ...